Coefficient of Restitution

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. I drop a ball from three point two metres, and it bounces back up to only one point eight metres. So, what fraction of its speed did the floor give back? Just divide the two heights, right? Hmm, not so fast. One number answers this exactly: the coefficient of restitution. It comes up often in JEE collision problems, so let's begin.

0:28Coefficient of restitution: separation over approach

So, in a collision, two bodies first come towards each other, and then move apart, right? The coefficient of restitution, the small letter e on screen, compares these two speeds: speed of separation over speed of approach. Okay? Now in symbols, for a head-on collision. u one and u two are the velocities before the impact, v one and v two after. The small one and two written low are subscripts, labels for the two bodies. So e equals v two minus v one, over u one minus u two, along the line of impact, with signs for directions. With me? Now think. Suppose the two bodies stick together after the impact. What is the value of e? Zero, yes, exactly! They never separate, so the speed of separation is zero. At the other end, e equals one is an elastic collision: they separate as fast as they approached. And zero less than e less than one, in between, is inelastic. The pointed signs mean less than. Clear? And one more point. e depends on the materials of the two bodies, not on their masses. Make one ball heavier, and e stays the same. Got it?

1:42Velocities after the collision

Now, the real use of e. After the collision we have two unknowns, v one and v two. So how many equations do we need, and where do they come from? Two equations, correct! First, momentum, conserved across the impact, with m one and m two the masses. m one u one plus m two u two equals m one v one plus m two v two. Okay? Second, cross multiply the definition of e. v two minus v one equals e times u one minus u two, in brackets. Two equations, two unknowns. That's it! Right? Solved in general, they give this. v one equals m one minus e m two, in brackets, times u one, plus one plus e, in brackets, times m two u two, all over m one plus m two. Clear? And v two has the same pattern, with one and two swapped. Quick check: put e equals zero, and both give the same answer, the common velocity of bodies that stick. Nice, isn't it? Now, kinetic energy. If e equals one, how much kinetic energy is lost? None at all, very good! Kinetic energy lost equals half, times mu, times one minus e squared, times u one minus u two, whole squared. Mu, the Greek letter, is the reduced mass, m one m two over m one plus m two. With e equal to one, one minus e squared is zero: no loss. With me?

3:17Ball on a fixed floor

Now, a case that comes up often: a ball dropped from height h onto a fixed floor. In the figure, the red dot is the ball at height h, each arch is one bounce, and the dashed lines mark the starting height and then how high the ball rises after each bounce. Okay? The ball hits the floor with speed v. The floor is fixed; it does not move. So what is the ball's speed just after the bounce? e times v, exactly! The floor stays at rest, so the speed of approach is v, and the speed of separation, the rebound speed, is e times v. Right? So if the speed becomes e times, the rebound height is also e times h. Simple, isn't it? Hmm, but look at the labels on the dashed lines: e squared h, then e to the power four, times h. The small two and four written up high are powers. So who is right? The figure is right! Oh, careful, e times h is a common trap. Height depends on the square of the speed. So if the speed becomes e times, the height becomes e squared times: rebound height e squared h. Clear? And every bounce repeats this. So after n bounces, the height is e to the power two n, times h. You can see it at the top right of the figure too. Okay? Now turn it around. Starting height h one, rebound height h two: e equals the square root of h two over h one. The tick shaped sign is the square root, and the slash means over. We need this in a minute! Got it?

4:54Bouncing to rest: total distance and time

For a ball dropped from height h, add up all the ups and downs until it comes to rest. The total distance is h times one plus e squared, over one minus e squared. It is on the right of the figure too. With me? And the total time to come to rest is the square root of two h over g, times one plus e, over one minus e, with g the acceleration due to gravity. Keep both handy for the exam. Right?

5:20Example 1: find e from drop and rebound heights

Alright, back to our puzzle, the first example from the course. Here, a ball dropped from three point two metres rebounds to one point eight metres. Find the coefficient of restitution, e, okay? The m stands for metres. Pause the video and try it. Done? Okay, let's check. Just divide the heights: e equals one point eight over three point two. Is that right? No, no! Height goes as e squared, remember? So e is the square root of h two over h one: root of one point eight over three point two. Okay? That is eighteen over thirty two, or nine over sixteen, and its root is three quarters. So e equals zero point seven five. Puzzle solved: the floor gave back three quarters of the ball's speed. Perfect!

6:12Example 2: velocities after a collision with e = 0.5

Now a harder one, again from the course. Here, a two kilogram ball at six metres per second hits a one kilogram ball at rest, with e equal to zero point five. Find the velocities after. k g is kilograms, and m s to the minus one is metres per second. Pause the video and try it. Done? Okay, let's check. Momentum first. Before: two times six, and the ball at rest adds nothing. After: two v one plus one times v two. So two v one plus v two equals twelve. Okay? Now restitution. What is the speed of approach here? Six minus zero, six, correct! So the speed of separation, v two minus v one, is zero point five times six: three. Careful with the order: v two minus v one, to match u one minus u two. Right? Now solve. v two equals v one plus three. Put it into the momentum equation: two v one plus v one plus three equals twelve, so three v one equals nine. With me? So v one is three metres per second, and v two is three plus three: six metres per second. Both move forward, and the light ball in front is faster, so they separate. Exactly as it should be! Quick check with the formula for v one. Two minus zero point five times one is one point five, times six is nine, plus zero, over two plus one. Nine over three: three again. Perfect!

7:48Quick revision: definition of e, velocities, rebound height

So, quick revision. e is speed of separation over speed of approach: one for elastic, zero when bodies stick. For velocities after a collision, momentum plus restitution: two equations. And on a fixed floor, speed becomes e times, height e squared times, so e is the root of h two over h one. Now go and try the practice problems on this page. All the best!